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  • A disc of mass and radius <img alt="\mathrm{\mathrm{R}}" src="/latex-image/?%5Cmathrm%7B%5Cmathrm%7

A disc of mass \mathrm{1 \mathrm{~kg}} and radius \mathrm{\mathrm{R}} is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest  position, its angular speed will be  \mathrm{4 \sqrt{\frac{x}{3 R}} \text { rad } \mathrm{s}^{-1}}  where \mathrm{ x=} ______________. \mathrm{\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)}

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

As the system is released and comes to the lowest position

\mathrm{T E_{\text {initial }} =T E_{\text {final }} }

\mathrm{M g h+0 =\frac{1}{2} I \omega^2 }

\mathrm{M g(2 R) =\frac{1}{2}\left(\frac{M R^2}{2}+M R^2\right) \omega^2 }

\mathrm{8 M g R =3 M R^2 \omega^2 }

\mathrm{\sqrt{\frac{8 g}{3 R}} =\omega }

\mathrm{16 x =8 g=80 }

\mathrm{x =5}

The value of \mathrm{x} is \mathrm{5}

Posted by

Rakesh

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